Novalee S.
asked • 10/28/24:(((((( Heloooppppp
Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If it diverges to infinity, state your answer as "oo" (without the quotation marks). If it diverges to negative infinity, state your answer as "-oo". If it diverges without being infinity or negative infinity, state your answer as "DNE".
∫ 13 on top and 7 on bottom (13)/(cubedroot(x-7))dx
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1 Expert Answer
Touba M. answered • 10/28/24
Tutor
4.9
(31)
B.S. in Pure Math with 20+ Years Teaching/Tutoring Experience
Hi,
∫ 13 on top and 7 on bottom (13)/(cubedroot(x-7))dx
your question is:
13 ∫ (x-7)^-1/3 dx As you know ∫ u^ndu = 1/(n+1) u^(u+1)
u= (x-7) n = -1/3 du = dx so the answer of integral will be
13 ( 3/2) (x-7)^2/3 7 to 13 =====> after replacing 7 and 13 you will have 39/2 ( 36)^1/3
so it is convergent and now you know the value of integral.
I hope it is useful.
Minoo
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Julius N.
To determine whether the integral\[ \int_{7}^{13} \frac{13}{\sqrt[3]{x -7}} \, dx\] is convergent or divergent, we start by identifying potential issues at the limits of integration. Specifically, since we are integrating from \(7\) to \(13\), we need to check what happens at \(x =7\) because the integrand may become problematic there. Step1: Behavior near the lower limitAs \(x\) approaches \(7\), the term \(\sqrt[3]{x -7}\) approaches \(0\), which would make the integrand \(\frac{13}{\sqrt[3]{x -7}}\) tend to \(\infty\). Therefore, we expect possible divergence at this limit. Step2: Determine the type of divergenceWe can rewrite the integral to better analyze it near \(x =7\): \[ \int_{7}^{13} \frac{13}{(x -7)^{1/3}} \, dx. \] Step3: Find the antiderivativeThe integral can be computed as follows: \[ \int \frac{13}{(x -7)^{1/3}} \, dx =13 \cdot \int (x -7)^{-1/3} \, dx. \] The antiderivative of \((x -7)^{-1/3}\) is: \[ \frac{(x -7)^{2/3}}{(2/3)} = \frac{3}{2}(x -7)^{2/3} + C. \] Thus, we have: \[ \int \frac{13}{(x -7)^{1/3}} \, dx =13 \cdot \frac{3}{2} (x -7)^{2/3} = \frac{39}{2} (x -7)^{2/3} + C. \] Step4: Evaluate the definite integralWe evaluate the integral from \(7\) to \(13\): \[ \left[ \frac{39}{2} (x -7)^{2/3} \right]_{7}^{13}. \] Calculating the upper limit: At \(x =13\): \[ \frac{39}{2} (13 -7)^{2/3} = \frac{39}{2} (6)^{2/3}. \] Now, calculating \((6)^{2/3}\): This is \((6^{2})^{1/3} =36^{1/3} =6^{2/3}\). At \(x =7\): \[ \lim_{x \to7^+} \frac{39}{2} (x -7)^{2/3} = \frac{39}{2}(0)^{2/3} =0. \] Step5: Evaluate the limitsPutting this together: \[ \left[ \frac{39}{2} (x -7)^{2/3} \right]_{7}^{13} = \frac{39}{2} \cdot6^{2/3} -0 = \frac{39}{2} \cdot6^{2/3}. \] Final ConclusionSince we are integrating from \(7\) to \(13\) and the integrand diverges negatively to \(-\infty\) as \(x\) approaches \(7\), the integral diverges. Thus, the final answer is: \[ \text{DNE}. \]10/29/24